Probability Questions for Uniform Random Variables on [0,1]

[e(ho0n3]

Homework Statement
Let {X_i} be a sequence of uniform random variables on [0,1]. What is P(sup {X_i} = 1) and P(X_i = X_j), i ≠ j?

The attempt at a solution
If sup{X_i} = 1, then either X_i is 1 for some i or there is an increasing sequence of the X_i's that converges to 1. The probability of the former is 0. I don't know how to calculate the probability of the latter.

For the second probability, if things were finite I would condition on the value of X_j and compute the probability that way. But in this case, I have no idea what to do.

 

[jbunniii]

Consider P(\sup\{x_i\} \neq 1). If the supremum is not 1, it must be less than one, say \sup\{x_i\} = \alpha < 1. If this is the case, then none of the x_i can be greater than \alpha. What is the probability of that?

For the second question, note that x_i = x_j if and only if x_i - x_j = 0.

 

[e(ho0n3]

Consider P(\sup\{x_i\} \neq 1). If the supremum is not 1, it must be less than one, say \sup\{x_i\} = \alpha < 1. If this is the case, then none of the x_i can be greater than \alpha. What is the probability of that?

Good tip. For a particular i, P(X_i > \alpha) = 1 - P(X_i \le \alpha) = 1 - \alpha, so the probability that X_i < \alpha for all i must be \prod_i (1 - \alpha) = 0. Hmm...this can't be right.

For the second question, note that x_i = x_j if and only if x_i - x_j = 0.

Right. I don't know why this didn't occur to me. Thanks.

 

[jbunniii]

Good tip. For a particular i, P(X_i > \alpha) = 1 - P(X_i \le \alpha) = 1 - \alpha, so the probability that X_i < \alpha for all i must be \prod_i (1 - \alpha) = 0. Hmm...this can't be right.

That's what I get. This shows that P(\sup\{x_i\} \leq \alpha) = 0 for all \alpha < 1. So P(\sup\{x_i\} = 1) must be 1.

It makes sense that this must be the case. If I choose infinitely many uniform numbers in [0,1], why should ALL of these numbers avoid some interval [\alpha,1] with nonzero probability? What's special about the rightmost edge of the interval? Nothing at all.

 

[e(ho0n3]

You're right. My intuition about such things is, as you can tell, not very good. Thanks.